10.01b - Speed, density and pressure | OCR GCSE Mathematics

Is travelling at $36\,\mathrm{km/h}$ faster than travelling at $12\,\mathrm{m/s}$ ? Understand what “per” means, use speed, pay, unit prices, density and pressure, and convert compound units correctly even when the quantities contain letters.

Speed and unit rates

Is $36\,\mathrm{km/h}$ faster than $12\,\mathrm{m/s}$ ? Comparing the numbers alone is unreliable: a rate connects two quantities, and both units matter. We will convert them to a common unit later.

A unit rate states an amount for one unit of another quantity. “ $60$ kilometres per hour” means $60$ kilometres in each hour at that constant speed. Dividing a journey's total distance by its total time gives its average speed, even when the speed varies during the journey.

Speed, distance and time

v=\frac{d}{t},\qquad d=vt,\qquad t=\frac{d}{v}

Use compatible units, with $t>0$ . The final rearrangement needs $v>0$ . These formulae are not on either 2027 formula sheet: learn their meaning.

If distance is in kilometres and time in hours, the quotient is in $\mathrm{km/h}$ . A time of $45$ minutes is $45/60=0.75$ hours, not $0.45$ hours. Include stops in the time if the question asks for the average over the whole journey.

Worked example 1: average speed

Question. A cyclist travels $18\,\mathrm{km}$ in $0.75$ hours. Calculate the average speed.

Plan. Divide distance by time in hours.

\begin{aligned} v&=18\div0.75\\ &=24\,\mathrm{km/h} \end{aligned}

Answer. $24\,\mathrm{km/h}$ .

Check. $24\times0.75=18$ kilometres, recovering the journey distance.

Change only the time notation. The next time describes the same duration.

Worked example 2: minutes supplied

Question. A cyclist travels $18\,\mathrm{km}$ in $45$ minutes. Calculate the average speed in $\mathrm{km/h}$ .

Plan. Convert minutes to hours before dividing.

\begin{aligned} t&=45\div60=0.75\,\mathrm{h}\\ v&=18\div0.75=24\,\mathrm{km/h} \end{aligned}

Answer. $24\,\mathrm{km/h}$ .

Check. $45$ minutes is three-quarters of an hour, so $18$ kilometres is three-quarters of $24$ kilometres.

Pay and prices use the same “amount divided by number of units” structure. If work is paid at a constant hourly rate, total pay equals rate multiplied by hours. A unit price lets you compare different pack sizes fairly.

Worked example 3: finding and using an hourly rate

Question. A worker earns $£42$ for $3.5$ hours at a constant rate. Find the hourly rate and the pay for $5$ hours.

Plan. Divide to find one hour's pay, then multiply for five hours.

\begin{aligned} \text{hourly rate}&=42\div3.5=£12\text{ per hour}\\ \text{five-hour pay}&=12\times5=£60 \end{aligned}

Answer. $£12$ per hour and $£60$ .

Check. Seven half-hours at $£6$ each give the original $£42$ .

Change only the duration for which the same pay was earned. A smaller number of hours gives a greater rate.

Worked example 4: a changed hourly rate

Question. A worker instead earns $£42$ for $3$ hours. At this constant rate, find the hourly rate and pay for $5$ hours.

Plan. Repeat the division and multiplication using the new duration.

\begin{aligned} \text{hourly rate}&=42\div3=£14\text{ per hour}\\ \text{five-hour pay}&=14\times5=£70 \end{aligned}

Answer. $£14$ per hour and $£70$ .

Check. $3\times14=42$ , and five hours earns more than three hours.

For unit pricing, first choose the same mass or volume unit for both products. The cheapest pack is not necessarily the cheapest per kilogram.

Worked example 5: comparing pack prices

Question. Rice costs $£1.60$ for $400\,\mathrm{g}$ , or $£3.60$ for $1\,\mathrm{kg}$ . Which has the lower price per kilogram?

Plan. Express both masses in kilograms, then divide cost by mass.

\begin{aligned} 400\,\mathrm{g}&=0.4\,\mathrm{kg}\\ \text{small-pack unit price}&=1.60\div0.4=£4\text{ per kg}\\ \text{large-pack unit price}&=3.60\div1=£3.60\text{ per kg} \end{aligned}

Answer. The $1\,\mathrm{kg}$ pack is cheaper per kilogram.

Check. Five small packs contain $2\,\mathrm{kg}$ and cost $£8$ . Two large packs contain the same mass and cost $£7.20$ .

Change only the small pack's price. Recalculate instead of assuming the larger pack always wins.

Worked example 6: a reduced pack price

Question. The $400\,\mathrm{g}$ pack now costs $£1.20$ . The $1\,\mathrm{kg}$ pack still costs $£3.60$ . Which has the lower unit price?

Plan. Divide the new small-pack price by $0.4$ kilograms.

\begin{aligned} \text{small-pack price}&=1.20\div0.4=£3\text{ per kg}\\ £3\text{ per kg}&<£3.60\text{ per kg} \end{aligned}

Answer. The $400\,\mathrm{g}$ pack is now cheaper per kilogram.

Check. Five small packs now cost $£6$ for $2\,\mathrm{kg}$ , below the $£7.20$ cost of two large packs.

A unit-price comparison assumes you want comparable products and can use the quantity bought. It measures price per unit, not every reason for choosing a product.

Watch out

Do not average two speeds unless the times spent at those speeds are equal. Average speed always uses total distance divided by total time.

Apply the same unit-price method to fresh packs.

Density and pressure

Two blocks can have the same volume but different masses. Density measures how much mass occupies one unit of volume. A density of $2.7\,\mathrm{g/cm^3}$ means each cubic centimetre has mass $2.7\,\mathrm{g}$ , assuming uniform material.

Density

\rho=\frac{m}{V},\qquad m=\rho V,\qquad V=\frac{m}{\rho}

Use compatible mass and volume units, with $V>0$ . The last rearrangement requires $\rho>0$ . The density relationship is not on the 2027 sheets: learn its meaning.

The symbol $\rho$ , pronounced “rho”, stands for density. You may also write the word density. A hollow object's overall density uses its total mass divided by its overall volume, which can differ from the density of the material alone.

Worked example 7: finding density

Question. A uniform block has mass $540\,\mathrm{g}$ and volume $200\,\mathrm{cm^3}$ . Calculate its density.

Plan. Divide mass by volume to find mass in one cubic centimetre.

\begin{aligned} \rho&=540\div200\\ &=2.7\,\mathrm{g/cm^3} \end{aligned}

Answer. $2.7\,\mathrm{g/cm^3}$ .

Check. $200\times2.7=540$ grams, recovering the mass.

Change only the mass of a block of the same volume. Its density changes in the same proportion.

Worked example 8: a heavier block

Question. A second uniform block has volume $200\,\mathrm{cm^3}$ and mass $780\,\mathrm{g}$ . Find its density.

Plan. Divide the new mass by the unchanged volume.

\begin{aligned} \rho&=780\div200\\ &=3.9\,\mathrm{g/cm^3} \end{aligned}

Answer. $3.9\,\mathrm{g/cm^3}$ .

Check. The mass rose by $240\,\mathrm{g}$ , so the density rose by $240/200=1.2\,\mathrm{g/cm^3}$ . Indeed, $3.9-2.7=1.2$ .

If density is $5\,\mathrm{g/cm^3}$ and volume is $12\,\mathrm{cm^3}$ , mass is $5\times12=60\,\mathrm{g}$ . Reversing this, a $60\,\mathrm{g}$ sample at that density occupies $60/5=12\,\mathrm{cm^3}$ . These follow from multiplying or dividing the mass-per-volume statement.

Pressure is force per unit contact area. A fixed force spread over a larger area gives smaller pressure. When using $P=F/A$ , $F$ is the perpendicular force and $A>0$ is the contact area. Force in newtons divided by area in square metres gives $\mathrm{N/m^2}$ , also called pascals, $\mathrm{Pa}$ . The pressure relationship is not on the 2027 sheets; use it when supplied or required by the question.

Worked example 9: pressure on one face

Question. A perpendicular force of $180\,\mathrm{N}$ acts over $12\,\mathrm{cm^2}$ . Calculate the pressure in $\mathrm{N/cm^2}$ , using $P=F/A$ .

Plan. Divide force by the stated contact area.

\begin{aligned} P&=180\div12\\ &=15\,\mathrm{N/cm^2} \end{aligned}

Answer. $15\,\mathrm{N/cm^2}$ .

Check. Twelve square centimetres, each carrying $15\,\mathrm{N}$ , account for all $180\,\mathrm{N}$ .

Change only the contact area. The same force is now spread more thinly.

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10.01b - Speed, density and pressure

Speed and unit rates

Density and pressure

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